$f(0)=2\,$, $~f\,^\prime(0)=8\,$, $~f\,^{\prime\prime}(0)=-3\,$, and $~f\,^{\prime\prime\prime}(0)=1\,$. What are the first four nonzero terms of the Maclaurin series of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $2x+8{{x}^{2}}-3{{x}^{3}}+{{x}^{4}}$ (Choice B) B $2+8x-\frac{3}{2}{{x}^{2}}+\frac{1}{6}{{x}^{3}}$ (Choice C) C $2+8x-3{{x}^{2}}+{{x}^{3}}$ (Choice D) D $2+8x-\frac{3}{2}{{x}^{2}}+\frac{1}{3}{{x}^{3}}$ (Choice E) E $2x+8{{x}^{2}}-\frac{3}{2}{{x}^{3}}+\frac{1}{3}{{x}^{4}}$
We know the formula for a Taylor series centered at $~x=0~$ is $ f(0)+f\,^\prime(0)x+\frac{f\,^{\prime\prime}(0)}{2!}{{x}^{2}}+\frac{f\,^{\prime\prime\prime}(0)}{3!}{{x}^{3}}+...+\frac{{{f}^{(n)}}(0)}{n!}{{x}^{n}}+...\,$ So, if we substitute with what is given for the first four terms, we get the following four-term Taylor polynomial. $ T_{3}(x)=2+8x-\frac{3}{2!}{{x}^{2}}+\frac{1}{3!}{{x}^{3}}$ This simplifies to $ T_{3}(x)=2+8x-\frac{3}{2}{{x}^{2}}+\frac{1}{6}{{x}^{3}}$.